Question

The time period of a simple pendulum is $T$. The time taken to complete $5/8$ oscillation starting from mean position is $\frac{\alpha }{12}T$. The value of $\alpha$ is _______.

Answer 1

Let the amplitude of oscillation is $A$, then the total distance covered during one oscillation is $4A$.

In $5/8$ oscillation, total distance covered will be $2.5A=2A+0.5A$.


From phasor diagram,
$\pi +\theta =\omega t$
$\pi +\frac{\pi }{6}=\left(\frac{2\pi }{T}\right)t$

Here, $\pi$ angle represents distance of $2A$ and $\frac{\pi }{6}$ angle represents distance of $0.5A$.

$\Rightarrow \frac{7\pi }{6}=\left(\frac{2\pi }{T}\right)t$

$\Rightarrow t=\frac{7}{12}T=\frac{\alpha }{12}T$

$\Rightarrow \alpha =7$